Question: $\int^{2}_{0}x{\sqrt{1+2x^2}}\,dx\, = $
Strategy Let's first find the indefinite integral $\int x{\sqrt{1+2x^2}}\,dx\,$. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int x{\sqrt{1+2x^2}}\,dx\,$, we can use U-substitution. If we let $ {u=1+2x^2}$, then ${du=4x \, dx}$ and ${ x\,dx=\dfrac{1}{4}\, du}$. So we have: ∫ x 1 + 2 x 2 − − − − − − √ d x = ∫ 1 + 2 x 2 − − − − − − √ ⋅ x d x = ∫ u √ ⋅ 1 4 d u = 1 4 ∫ u √ d u = 1 4 ⋅ 2 3 u 3 2 + C = 1 6 u 3 2 + C = 1 6 ( 1 + 2 x 2 ) 3 2 + C \begin{aligned}\int x{\sqrt{1+2x^2}}\,dx\,&=\int {\sqrt{{1+2x^2}}}\,\cdot {x\,dx}\,\\\\\\\\ &=\int\sqrt{ u}\,\cdot {\dfrac14\, du}\,\\\\\\\\ &=\dfrac14\int\sqrt{u}\,du\\\\\\\\ &=\dfrac{1}{4}\cdot \dfrac23u\^{\frac32}+C\\\\\\\\ &=\dfrac16u\^{\frac32}+C\\\\\\\\ &=\dfrac16\left(1+2x^2\right)\^{\frac32}+C \end{aligned} Evaluating the definite integral Now let's compute the definite integral: ∫ 2 0 x 1 + 2 x 2 − − − − − − √ d x = 1 6 ( 1 + 2 x 2 ) 3 2 ∣ ∣ ∣ 2 0 = 1 6 ( 27 − 1 ) = 26 6 = 13 3 \begin{aligned}\int^{2}_{0}x{\sqrt{1+2x^2}}\,dx\, &=\dfrac16\left(1+2x^2\right)\^{\frac32}\Bigg|^2_0\\\\\\\\ &=\dfrac{1}{6}\left(27-1\right)\\\\\\\\ &=\dfrac{26}{6}\\\\\\\\ &=\dfrac{13}{3}\end{aligned} [Did we have to find the indefinite integral first?] The answer $\int^{2}_{0}x{\sqrt{1+2x^2}}\,dx\, = \dfrac{13}3$